Question

The $x$ and $y$ co-ordinates of a particle at any time are $x=5t^2-3t+6$ and $y=10t^2+2t+3$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. Find the acceleration of the particle at $t=1\text{s}$.

• A. $(10\hat{i}+10\hat{j}){\text{m/s}}^2$
• B. $(10\hat{i}+20\hat{j}){\text{m/s}}^2$
• C. $(20\hat{i}+10\hat{j}){\text{m/s}}^2$
• D. $(10\hat{i}-20\hat{j}){\text{m/s}}^2$

Answer 1

The correct option is B $(10\hat{i}+20\hat{j}){\text{m/s}}^2$

Given,
$x$ co-ordinate of the particle: $x=5t^2-3t+6$
$y$ co-ordinate of the particle:$y=10t^2+2t+3$
To find: acceleration $a$ at $t=1\text{s}$
We know that,
Velocity, $v=\frac{dr}{dt}[r\to \text{displacement}]$
Acceleration , $a=\frac{dv}{dt}$
Along $x$ - axis
$v_x=\frac{dx}{dt}=(10t-3)\text{m/s}$
$\&a_x=\frac{d{v}_x}{dt}=10{\text{m/s}}^2$
Along $y$ - axis -
$v_y=\frac{dy}{dt}=(20t+2)\text{m/s}$
$\&a_y=\frac{d{v}_y}{dt}=20{\text{m/s}}^2$
$\therefore \overrightarrow{a}=a_x\hat{i}+a_y\hat{j}=(10\hat{i}+20\hat{j}){\text{m/s}}^2$